Short exact sequence of homotopy groups

Question

Let $p :E \to B$ be a weak fibration with section $s:B \to E$ and $F=p^{-1}(b_0)$. Show that the sequence $$0\longrightarrow \pi_n(F,s(b_0)) \xrightarrow{\quad i_*\quad}\pi_n(E,s(b_0))\xrightarrow{\quad p_*\quad}\pi_n(B,b_0)\longrightarrow 0$$ is exact at $\pi_n(E,s(b_0))$ for $n\ge1$.

To prove that $\operatorname{im}i_* \subset \ker p_*$ we can consider $[\alpha]=i_*([\beta])$ for $[\beta] \in \pi_n(F,s(b_0))$.

Now $$p_*([\alpha]) = [p\circ i \circ \beta] = [c_{b_0}]$$ i.e. $p_*([\alpha])$ is the identity.

I'm having some trouble proving that $\ker p_*\subset \operatorname{im}i_* $.

If we pick $[\alpha]\in \ker p_*$, then $\alpha : (I^n,\partial I^n) \to (E,s(b_0))$ and $p_*([\alpha])=[c_{b_0}]$. Now I need to construct an element $[\beta] \in \pi_n(F,s(b_0))$ such that $[\alpha]=i_*([\beta])$, but I have no idea how to come up with such $[\beta]$. What properties of the homotopy groups could I use here?

Answer 1

Since $p:E\to B$ is a weak fibration, it has homotopy lifting property with respect to all CW complexes. This is not immediately applicable as we are working with homotopy relative to subspace which requires additional constraints, but we can use the following result derived from HLP:

If $p:E\to B$ is a weak fibration, then it satisfies the relative homotopy lifting property with respect to the pair $(I^n,\partial I^n)$.

The idea of relative homotopy lifting with respect to the pair $(X,A)$ is as follows. Suppose we are given a homotopy $H$ between $f,g:X\to B$, a lifting $\tilde f$ s.t. $p\circ \tilde f=f$, and a lifted homotopy $\tilde h$ defined on $A\subseteq X$ such that $\tilde h|_{A\times\{0\}}=\tilde f|_{A}$, i.e., $p\circ \tilde h=H|_{A\times I}$, then we can find a lifting $\tilde H:X\times I\to E$ s.t. $\tilde H_{X\times\{0\}}=\tilde f$ and $\tilde H|_{A\times I}=\tilde h$. In terms of diagram (It seems that it's still impossible to draw diagonal arrows), this means that there exists $\tilde H:X\times I\to E$ that fills the diagram below. (Just imagine there is a map $\tilde H$ connecting the diagonal). $\require{AMScd}$ \begin{CD} X\times\{0\}\cup A\times I @>{\tilde{f}\cup\ \tilde h}>> E\\ @V{i}VV @VV{p}V\\ X\times I @>{H}>> B \end{CD} The diagram of (the ordinary version of) HLP is essentially the special case $A=\varnothing$.

Continue from your argument. We are given $\alpha:I^n\to E$ with $\alpha(\partial I^n)=\{s(b_0)\}$ and a homotopy $H:I^n\times I\to B$ relative to $\partial I^n$ between $p\circ \alpha$ and $c_{b_0}$. Restricting to $\partial I^n\times I$, $H$ can be lifted to $\tilde h=c_{s(b_0)}$. By relative homotopy lifting property, we can find a homotopy $\tilde H:I^n\times I\to E$ such that $p\circ \tilde H=H$ and $\tilde H|_{I^n\times\{0\}}=\alpha$, and $\tilde{H}|_{\partial I^n\times I}=c_{s(b_0)}$, so $\tilde H$ gives a homotopy between $\alpha$ and $\gamma=\tilde H|_{I^n\times\{1\}}$ relative to $\partial I^n$.

Now, we have $p\circ \gamma=p\circ \tilde H|_{I^n\times\{1\}}=c_{b_0}$, meaning that $\gamma(I^n)\subseteq p^{-1}(b_0)=F$. Therefore $\gamma$ factors through $F=p^{-1}(b_0)$, that is, we can find a map $\beta:(I^n,\partial I^n)\to (F,s(b_0))$ such that $\gamma=i\circ \beta$, where $i:(F,s(b_0))\hookrightarrow (E,s(b_0))$ is the inclusion of fiber. This shows that $i_\ast([\beta])=[\gamma]=[\alpha]$, since $\alpha\simeq\gamma$ rel $\partial I^n$.

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For a detailed proof of relative homotopy lifting property with respect to $(I^n,\partial I^n)$, see here. The proof uses a special homeomorphism of pairs $h:(I^n\times I, I^n\times\{0\}\cup\partial I^n\times I)\to (I^n\times I,I^n\times \{0\})$. For more information of this map, see here and here.